1. Slide 1 5. VHDL/Verilog Operators, Arrays and Indexes

2. Slide 2 Verilog for synthesis: What operators can we use? Operators can be used in both assign and always statements Logical bitwise operators: &, |, ~, ^, ~^ Operation applies to each bit of a signal Therefore, apply operators to signals with the SAME WIDTH only Otherwise, the signal with lower width will be left aligned (some synthesizers might align it to right) Logical operators: !, &&, ||, ==, !=, ===,, =, !== Used in conditions only. Same as in C, a condition is written always in () What is === and !== ? Signals are compared including X and Z Note: X = don’t care, not undefined! Applies to simulation only!

3. Slide 3 Verilog for synthesis: What operators can we use? Operators can be used in both assign and always statements Arithmetic operators: +, -, *, /, %, ** /, % and ** can be synthesized only if the second operand is 2 or power of 2! Shift: >, >> Shifted in values are 0 >> : shift and maintain sign bit Unary reduction operators &, ~&, |, ~|, ~, ^, ~^, Concatenate and replicate { }, {{}}

4. Slide 4 Verilog for synthesis: What operators can we use? Unary reduction operators Example: We have a N-bit wide signal named A. The width of A is determined by a parameter called SIGNAL_WIDTH. We have to make a logic AND between all of the bits of A into a signal called A_and. How can we do this? Solution 1: … wire [SIGNAL_WIDTH-1:0] A; reg A_and; integer I; always@ (A) for (i=0; i<SIGNAL_WIDTH; i = i+1) A_and <= A_and & A[i]; Solution 2: … wire A_and; assign A_and = &A; Note: To test whether any of the bits of A is 1, one can use (|A)

5. Slide 5 Verilog for synthesis: What operators can we use? Concatenate and replicate Concatenate Examples: 1. {a, b, c} – Concatenate a, b, c into a bus The width of the resulting bus = width of a + width of b + width of c Concatenate assignment is also positional! The leftmost bit of the resulting bus is the leftmost bit of a 2. Assume the following code snippet: wire [7:0] a; wire [2:0] b; wire [4:0] c; wire [18:0] q; assign q = {a[5:0], c[3:0], b, a[7:6], 1’b0, 2’b10, b[2]}; What will be connected to q[17], q[7] and q[2]?

6. Slide 6 Verilog for synthesis: What operators can we use? Concatenate and replicate Replicate Examples: 1.We have a N-bit wide signal called A and a 1-bit wide signal called EN. The value of N is determined by the parameter SIGNAL_WIDTH. The EN signal acts as an enable for the N-bit wide output signal A_out such as A_out = A if EN = 1, otherwise all of the bits of A are 0. How can we do this? Solution 1: assign A_out = (EN)? A:0; //works only if A is less than 32-bit wide! Otherwise: N’h00…. Solution 2: assign A_out = A & {SIGNAL_WIDTH{EN}}; 2. wire [19:0] q = {{2{4’hA}}, {4{3’b101}}}. Write down in hexadecimal the value of Q!

7. Slide 7 Verilog for synthesis: About indexes One-dimension signals i.e. BUSES: How to access a part of a bus? Example: wire | reg [N:n] A; N and n are POSITIVE INTEGER numbers A[Upper:Lower] always represents the CONTINUOUS part of the bus having the width of Upper-Lower+1 Assume Upper >= Lower, then ALWAYS Upper =n! Otherwise the Verilog parser generates an error. Careful when using parameters or variables for indexing! The synthesizer is able to determine if the index is out of the range [N:n]. Example: reg [15:0] A; … for (i=0; i<=16; i++) A[i] <= … //will generate an error For selecting non-continuous parts of A, use concatenation operators A[i] ALWAYS represents a 1-bit wide signal. Obviously, n<=i<=N A ALWAYS represents the WHOLE range of A, equivalent to A[N:n] Note: A[N:N] is valid if A was defined in this way! Example: parameter SIGNAL_WIDTH = 0; reg [SIGNAL_WIDTH-1:0] A; //A will be defined as A[0:0]

8. Slide 8 Verilog for synthesis: About indexes One-dimension signals i.e. BUSES: How to access a part of a bus? Example: wire | reg [N:n] A; What about accessing part of A as A[Lower:Upper]? Syntactically it is accepted but the synthesizer will generate an error Use the endianness of signals in the way they wer defined! RECCOMENDATION: use only one endianness in your code! (usually little) If cannot avoid both little and big endian signals, in your code, Example: wire [Lower:Upper] B = A [Upper:Lower];//B[Lower] = A[Upper]… Two-dimension signals i.e. ARRAYS Example: reg [N:n] A [m:M]; //It can be also [M:m] Used mostly for MEMORY structures: RAM, ROM, FIFO and multi- dimensional shift registers ALWAYS THE SECOND INDEX COMES FIRST! A[ i ], m<= i <=M ALWAYS represents a N-n+1-width signal i.e the i-th location of the memory or shift register

9. Slide 9 Verilog for synthesis: About indexes Two-dimension signals i.e. ARRAYS Example: reg [N:n] A [m:M]; //It can be also [M:m] A[ i ], m<=i<=M ALWAYS represents a N-n+1-width signal i.e the i-th location of the memory or shift register, equivalent to A [ i ] [N:n] A[ i ][ j ], m<= i <=M, n <= i <= N ALWAYS represents bit j of location i A[ i ][Upper:Lower] ALWAYS represents a part selector of the bus from location i A [L1:U1] [U2:L2] is syntactically valid, but the synthesizer will generate an error For example: reg [7:0] Mem [0:255]; … always @.. begin Mem <= 0; //NOT ALLOWED. Mem is an array of 256X8! Not 32 bits //and not even 2048 bits! Mem [0:3] <= {32{1’b0}}; //Also not allowed Mem [3] <= 8’hFF;//Allowed Mem [255] [7:4] <= 4’h8;//Allowed

10. Slide 10 Verilog for synthesis: About indexes Multi-Dimension signals? Example: reg [N:n] A [m:M] [p:P]; //It can be also [m:M], [P:p] and so on It will be visible as a three-dimensional memory ALWAYS, THE SECOND INDEX COMES FIRST! XST solves the three-dimensional array as a two-dimensional array of (P-p+1) * (N-n+1) X (M-m+1) The third index i.e. [p:P] represents a multiplier for the first index i.e. [N:n] Therefore A[ i ] [ j ], m<= i <=M, p <= j <= P represents a N-n+1-width signal i.e the i-th location, bits [ (( j+1 ) * N) + j : (( j+1 ) * N ) + j – (N-n) ] of the memory, or shift register, equivalent to A [ i ] [ j ] [N:n] A[ i ][ j ][ k ], m<= i <=M, p <= i <= P, n <= k <=N represents bit j*k of location i A[ i ][ j ][Upper:Lower] represents a part selector [ (j * Upper: j * Lower] of the bus from location i A [L1:U1] [U2:L2] is not accepted by the synthesizer, it will generate an error Example: reg [7:0] Mem [0:255][0:3]; //will generate an array of 256 X 32- bit wide registers

11. Slide 11 VHDL: About indexes One-dimension signals i.e. BUSES: Example: signal A: std_logic_vector (7 downto 0); signal B, C: std_logic_vector (0 to 3); Index can be also negative! Part selectors: Valid: A (5 downto 4); B(0 to 2); A (5 downto 5); C(0 to 3); Invalid, even syntactically is not accepted to change the endianness or using indexes larger than the range: A (3 to 7); B(3 downto 0); C(0 downto 0); A(8), A(6 downto -1); A is equivalent to A(7 downto 0), B to B(0 to 3) and so on 1-bit value: enclosed in ‘ ‘, multiple-bit values: enclosed in “ “ Hexadecimal prefix: X Concatenation operator: &. Examples: A <= C & B; -- A(7) <= C(0), A(6) <= C(1) and so on A <= ‘0’ & B &”000”; A<= X”FF”; B <= C(0 to 2) & ‘1’;

12. Slide 12 VHDL: About indexes Multi-dimension signals i.e. ARRAYS: First, the type of array has to be defined. Example: type my_mem2type is array (0 to 7) of std_logic_vector ( 15 downto 0); type my_mem3type is array (0 to 3) of my_mem2type; signal A: my_mem2type; signal B: my_mem3type; The first defined index comes first! Part selectors: similar to Verilog A(7); A(5) (5 downto 4); B(3) (7) (15); Invalid: A (3 to 7); B(2 to 3) etc.